A Daniel cell is made at 25,^oC by connecting | vedclass

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(A) The cell reaction is $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.$E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.34 - (-0.76) = 1.10\ V$.A

Vedclass · Accepted Answer

Concentration of $Zn^{2+}$ becomes $1\ M$ when $emf$ reaches $1.10\ V$.

Vedclass · Answer

(A) The cell reaction is $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.$E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.34 - (-0.76) = 1.10\ V$.At $emf = 1.10\ V$,$E_{cell} = E^0_{cell} - \frac{0.06}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} = 1.10\ V$.This implies $\log \frac{[Zn^{2+}]}{[Cu^{2+}]} = 0$,so $[Zn^{2+}] = [Cu^{2+}]$.Let $x$ be the amount of $Zn$ oxidized. $[Zn^{2+}] = 0.1 + x$ and $[Cu^{2+}] = 0.9 - x$.$0.1 + x = 0.9 - x$ $\Rightarrow 2x = 0.8$ $\Rightarrow x = 0.4$.So,$[Zn^{2+}] = 0.5\ M$ and $[Cu^{2+}] = 0.5\ M$. Option $A$ is incorrect as $[Zn^{2+}]$ becomes $0.5\ M$,not $1\ M$.After charging with $0.6\ F$,$0.3\ mol$ of $Zn^{2+}$ is reduced to $Zn$ and $0.3\ mol$ of $Cu$ is oxidized to $Cu^{2+}$.New $[Zn^{2+}] = 0.5 - 0.3 = 0.2\ M$ and $[Cu^{2+}] = 0.5 + 0.3 = 0.8\ M$.$E_{cell} = 1.10 - 0.03 \log \frac{0.2}{0.8} = 1.10 - 0.03 \log(0.25) = 1.10 - 0.03(-0.6) = 1.10 + 0.018 = 1.118\ V$.

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